Double Integral volumes: Triangular base

In summary, the problem is to evaluate the integral of 5x^2 + 2y^2 over the triangle with vertices at (1,1), (2,0), and (2,2). The correct setup for the limits of integration is y=x to y=2-x for y and 2 to 1 for x. However, there may be an error in the integration process as the final answer of 33/2 is not being reached. Further clarification or assistance is needed to determine the error.
  • #1
dan5
9
0
Evaluate

∫∫R 5x2 + 2y2

where R is triangle (1,1) (2,0) (2,2)

I see the lines bounding the triangle are y=x y=2-x and x=2, and have tried many attempts at setting up the correct limits.

Would it be correct to split this into 2 triangles, or are the limits y=x∫y=2-x for y and 2∫1 for x, correct
 
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  • #2
y=x∫y=2-x for y and 2∫1 for x

This.
 
  • #3
Thanks for the help... but

when I then integrate WRT y I get

2∫1 [5x2y+(2/3) y3] which I have to substitute in my limits of y=x and y=2-x

but then this leaves me with a (2/3) (2-x)3 term, which does not simplify well, and does not lead me to the correct answer.

Any ideas where I have gone wrong, thanks!
 
  • #4
Well I would just leave that term in the form (2/3) (2-x)^3 and then integrate it with respect to x without factoring it out, that term should just become (-1/6) (2-x)^4, by the power rule. Beyond that I can't see where you went wrong without seeing your work.
 
  • #5
Right, here goes...

Integrating WRT y

2∫1 [5x2y+(2/3)y3 ]

then entering limits

2∫1 [5x3+(2/3)x3] - [5x2(2-x) + (2/3)(2-x)3]

simplifying to...

2∫1 (32/3)x3 - 10x2 - (2/3)(2-x)3

applying the second integral WRT x

[ (32/12) x4 - (10/3) x3 + (1/6)(2-x)4 ]

then entering limits of 2 and 1 leaves me with 14, where I am meant to find 33/2.

Again thanks in advance! I can't help thinking my original set up of the limits in the problem is wrong
 
  • #6
hmm, I keep getting 75/2, I'm quite confident the limits of integration are correct, I'm not sure what is going wrong..
 

1. What is a double integral volume with a triangular base?

A double integral volume with a triangular base is a type of mathematical calculation used in calculus and geometry to find the volume of a three-dimensional shape with a triangular base. It involves finding the volume of the shape by integrating over two variables, typically x and y, within a certain region.

2. How is a double integral volume with a triangular base calculated?

To calculate a double integral volume with a triangular base, you first need to define the region of integration, which in this case is the triangular base. Then, you integrate the function over the two variables (x and y) within that region to find the volume. This can be done using various methods such as the disk or washer method, or by using polar coordinates.

3. What types of shapes can be represented with a double integral volume with a triangular base?

A double integral volume with a triangular base can represent a variety of three-dimensional shapes, including pyramids, cones, and triangular prisms. These shapes all have a triangular base and can be defined by a function in terms of x and y.

4. Are there real-world applications for double integral volumes with a triangular base?

Yes, there are many real-world applications for double integral volumes with a triangular base. They can be used in engineering and architecture to calculate the volume of structures with triangular bases, such as roofs or tanks. They are also used in physics to calculate the volume of irregularly shaped objects, such as rocks or bones.

5. What is the importance of understanding double integral volumes with a triangular base?

Understanding double integral volumes with a triangular base is important for anyone working in fields that involve three-dimensional calculations and measurements. It is also a fundamental concept in calculus and can help in solving more complex problems involving integration and volume calculations.

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